Rationalise the Denominator: Watch the Conjugate Flush the Irrational Upstairs

A fraction like \dfrac{1}{1 + \sqrt{3}} has a perfectly valid value — roughly 0.366 — but the irrational sitting in the denominator makes it awkward to work with. You cannot add it cleanly to another fraction, and the shape of the expression hides how big it actually is.

The fix is almost magical when you see it the first time. Multiply top and bottom by the conjugate of the denominator — the same two terms, with the middle sign flipped — and the irrational on the bottom vanishes. The radical does not disappear; it simply slides upstairs into the numerator, where it is much easier to read, compare, and combine with other expressions.

This page is a visual walkthrough of exactly how that happens.

What the conjugate is

If the denominator is a + b\sqrt{c}, its conjugate is a - b\sqrt{c} — the same two terms, with the sign between them swapped. If the denominator is \sqrt{m} - \sqrt{n}, the conjugate is \sqrt{m} + \sqrt{n}. The rule is always "flip the middle sign, leave everything else alone."

The reason the conjugate is useful is the difference-of-squares identity you already know from algebra:

(a + b)(a - b) = a^2 - b^2

When you multiply a binomial by its conjugate, the two cross terms cancel and you are left with just the squares. The squares of radicals are rational (because (\sqrt{c})^2 = c by definition of the square root). So the conjugate product is always rational — exactly what you want for a clean denominator.

Interactive: watch the irrational flush out

The slider below sweeps a parameter a through the expression \dfrac{1}{a + \sqrt{3}}. Drag the red dot to change a from 1 up to 5. The upper bar shows the original fraction; the lower bar shows the rationalised form \dfrac{a - \sqrt{3}}{a^2 - 3} produced by multiplying top and bottom by the conjugate a - \sqrt{3}. Notice that both bars have the same numerical value — only the shape of the expression is changing.

Drag the red point to change $a$ in the expression $\dfrac{1}{a + \sqrt{3}}$. The top orange bar shows the value before rationalisation; the bottom green bar shows the value of the rationalised form $\dfrac{a - \sqrt{3}}{a^2 - 3}$. Both bars stay the same length at every slider position — because rationalising multiplies by $1$, the *value* is unchanged. Only the *shape* of the expression moves the radical from the denominator to the numerator.

At a = 2, the original is \dfrac{1}{2 + \sqrt{3}} \approx 0.268, and the rationalised form is \dfrac{2 - \sqrt{3}}{2^2 - 3} = \dfrac{2 - \sqrt{3}}{1} = 2 - \sqrt{3} \approx 0.268. Same number, much friendlier form.

The three steps, with the algebra exposed

Pick a = 2 to make the working concrete. The original fraction is

\frac{1}{2 + \sqrt{3}}

Step 1. Write the conjugate. The denominator is 2 + \sqrt{3}, so its conjugate is 2 - \sqrt{3}. Flip the middle sign, leave the two terms alone.

Step 2. Multiply top and bottom by the conjugate. Multiplying a fraction by \frac{\text{thing}}{\text{thing}} is multiplying by 1, so the value is unchanged:

\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})}

Why: \frac{2-\sqrt{3}}{2-\sqrt{3}} = 1, so you have multiplied by 1. The number hasn't changed — the only thing that changed is the shape, because the denominator is about to be simplified by the difference-of-squares identity.

Step 3. Expand the denominator with (a + b)(a - b) = a^2 - b^2.

(2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1

The cross terms 2 \cdot (-\sqrt{3}) + \sqrt{3} \cdot 2 = 0 cancel exactly, leaving only the two squared terms. Because \sqrt{3} squared is the rational integer 3, the whole denominator collapses to 4 - 3 = 1.

\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}

The radical is now in the numerator, and the denominator is the clean integer 1. Done.

The same fraction in three forms. The middle form shows the cross-terms about to cancel; the right form is the result after the difference-of-squares collapses the denominator.

Why the conjugate always works

The conjugate trick is not a lucky coincidence — it is a direct consequence of one algebraic identity. Any binomial of the form a + b\sqrt{c} multiplied by a - b\sqrt{c} gives

(a + b\sqrt{c})(a - b\sqrt{c}) = a^2 - b^2 c

Every term on the right is rational (because a, b, c are rational), and there are no radicals left. The only way to combine two binomials involving \sqrt{c} so that the result is rational is to use the conjugate — any other pairing leaves a \sqrt{c} survivor in the product.

This is why the move is so reliable: for every denominator of the form a + b\sqrt{c} (or with two different radicals, \sqrt{m} \pm \sqrt{n}), there is exactly one multiplier that flattens the denominator into a rational integer, and that multiplier is the conjugate.

Two different radicals: the trick still works

What if the denominator has two different radicals, like \sqrt{5} - \sqrt{3}? The conjugate rule is still "flip the middle sign": the conjugate is \sqrt{5} + \sqrt{3}. Their product is

(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3}) = (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2

A clean integer, again with no radicals. So \dfrac{1}{\sqrt{5} - \sqrt{3}} = \dfrac{\sqrt{5} + \sqrt{3}}{2}, which is the standard simplified form.

The two-radical case is the one you will meet most often in Quadratic Equations and in the simplification of answers from the quadratic formula, where denominators like \sqrt{b^2 - 4ac} \pm \text{something} are the rule, not the exception.

When to use the simple version instead

If the denominator is a single radical like \sqrt{3} alone (not a binomial), you do not need the conjugate. Just multiply top and bottom by the same radical:

\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}

This is the one-term version of the same idea — the "conjugate" of a single term \sqrt{3} is itself, because there is no sign to flip. The denominator becomes (\sqrt{3})^2 = 3, and the radical moves upstairs.

Use the conjugate version only when the denominator is a binomial; for a single radical, the simpler one-term trick is enough.

Why rationalising matters

Rationalising is not cosmetic. Three real reasons to bother:

Comparison is easier. Is \dfrac{1}{1 + \sqrt{3}} bigger or smaller than \dfrac{1}{\sqrt{5}}? With radicals in both denominators, you cannot tell at a glance. Rationalise both and compare \dfrac{\sqrt{3} - 1}{2} with \dfrac{\sqrt{5}}{5}; the rational denominators make numerical estimation straightforward.
Adding fractions becomes possible. Two fractions with radical denominators cannot be added directly until they share a common denominator, and finding the LCM of \sqrt{3} and \sqrt{2} is meaningless. Rationalise first, then add as ordinary fractions.
Answer-key matching. Almost every school and JEE answer key expresses final answers with rational denominators. If your answer has a radical in the denominator, you will be marked down even when the value is correct. Rationalising is the last step of almost every problem involving roots.

The takeaway

Multiplying top and bottom by the conjugate uses the difference-of-squares identity to collapse an irrational binomial denominator into a rational integer. The value of the fraction doesn't change; only the form does. Every time you see a fraction with a binomial radical in the denominator, the reflex is instant: flip the middle sign, multiply by that conjugate over itself, watch the cross terms cancel.